Question

part b
how many moles of $o_{2}$ will be in the reaction vessel once the reactants have reacted as much as possible? (assume 100\\% actual yield)
express your answer using two significant figures.
$n =$

Explanation:

Step1: Analyze the reaction equation

The balanced reaction is \(2S(s)+3O_{2}(g)
ightarrow 2SO_{3}(g)\). The mole ratio of \(S\) to \(O_{2}\) is \(2:3\).

Step2: Determine the moles of \(O_{2}\) consumed

From the problem, we know that \(9\) moles of \(O_{2}\) react. Using the mole ratio from the balanced equation \(\frac{2\ mol\ S}{3\ mol\ O_{2}}\), the moles of \(S\) consumed is \(9\ mol\ O_{2}\times\frac{2\ mol\ S}{3\ mol\ O_{2}} = 6\ mol\ S\).

Step3: Calculate the excess \(S\)

The initial moles of \(S\) is \(7\ mol\). The excess \(S\) is \(7\ mol\ S-6\ mol\ S = 1\ mol\ S\).

Step4: Find moles of \(O_{2}\) remaining

We use the mole ratio \(\frac{3\ mol\ O_{2}}{2\ mol\ S}\). Since \(1\ mol\ S\) is in excess, the moles of \(O_{2}\) that would react with this excess \(S\) is \(1\ mol\ S\times\frac{3\ mol\ O_{2}}{2\ mol\ S}=1.5\ mol\ O_{2}\).

Step5: Calculate moles of \(O_{2}\) in the vessel

If we assume the initial moles of \(O_{2}\) is \(9\ mol\), the moles of \(O_{2}\) remaining in the vessel is \(1.5\ mol\) (rounded to two - significant figures).

Answer:

\(1.5\ mol\)